A uniform rod of length 200 cm and mass 500 g balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 A and another unknown mass 'm' is suspended the rod at 160 cm mark as shown in the Find the value of ' m ' such that the
rod is in equilibrium. (g = 10 m/s²)
A uniform rod of length 200 cm and mass 500 g balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 A and another unknown mass 'm' is suspended the rod at 160 cm mark as shown in the Find the value of ' m ' such that the
rod is in equilibrium. (g = 10 m/s²)
Option 1 - <p>1/6 kg</p>
Option 2 - <p>1/3 kg<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>1/2 kg<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>1/12 kg</p>
4 Views|Posted 5 months ago
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A
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5 months ago
Correct Option - 2
Detailed Solution:
Similar Questions for you
From A to B the process is isobaric

= W = 2 × R (600 - 350)
= 500 R
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